The correct answer is (D):
Let the area of ABCD be 100. Side of ABCD = 10 Area of EFGH is 62.5 => Side of EFGH = √62.5
Triangles AEH, BFE, CGF and DHG are congruent by ASA.
Let AE = BF = CG = DH = x; EB = FC = DG = AH = 10 -xx
AE2+ AH2 = EH2
x2+(10-x)2=(√62.5)2
Solving, x = 2.5 or 7.5
Since it’s given that CG is longer than EB, CG = 7.5 and EB = 2.5.
Therefore, EB : CG = 1 : 3
In the given figure, the numbers associated with the rectangle, triangle, and ellipse are 1, 2, and 3, respectively. Which one among the given options is the most appropriate combination of \( P \), \( Q \), and \( R \)?

The center of a circle $ C $ is at the center of the ellipse $ E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, where $ a>b $. Let $ C $ pass through the foci $ F_1 $ and $ F_2 $ of $ E $ such that the circle $ C $ and the ellipse $ E $ intersect at four points. Let $ P $ be one of these four points. If the area of the triangle $ PF_1F_2 $ is 30 and the length of the major axis of $ E $ is 17, then the distance between the foci of $ E $ is: